if-tan-5-tan-4-1-find-3-

Question Number 61157 by Askash last updated on 29/May/19

i f

t a n 5 θ + t a n 4 θ = 1

f i n d 3 θ

Commented by mr W last updated on 30/May/19

t h e r e a r e t o t a l l y 9 v a l u e s f o r \tan 3 θ .

o n e o f t h e m i s θ = \frac{π}{4}, a n d \tan 3 θ = - 1 .

w i t h θ = \frac{π}{4}

\tan 3 θ = \tan (\frac{3 π}{4}) = - 1

\tan 5 θ + \tan 4 θ = \tan (\frac{5 π}{4}) + \tan (π) = 1

Commented by Askash last updated on 29/May/19

t a n 3 θ

Commented by mr W last updated on 30/May/19

Answered by tanmay last updated on 30/May/19

t a n (α + β + γ + \dots) = \frac{S_{1} - S_{3} + S_{5} \dots}{1 - S_{2} + S_{4} - S_{6} + . .}

t a n θ = a

t a n 5 θ = \frac{5 a - 5 C_{3} a^{3} + 5 C_{5} a^{5}}{1 - 5 C_{2} a^{2} + 5 C_{4} a^{4}} = \frac{5 a - 10 a^{3} + a^{5}}{1 - 10 a^{2} + 5 a^{4}}

t a n 4 θ = \frac{4 a - 4 C_{3} a^{3}}{1 - 4 C_{2} a^{2} + 4 C_{4} a^{4}} = \frac{4 a - 4 a^{3}}{1 - 6 a^{2} + a^{4}}

t a n 5 θ + t a 4 θ = 1

\frac{5 a - 10 a^{2} + a^{5}}{1 - 10 a^{2} + 5 a^{4}} + \frac{4 a - 4 a^{3}}{1 - 6 a^{2} + a^{4}} = 1

w a i t \dots

a n i t h e r a p p r o a c h \dots

h e r e t a n 5 θ + t a n 4 θ = 1

s o θ = s m a l l

t a n 5 θ + t a n 4 θ \approx 5 θ + 4 θ

9 θ = 1

θ = \frac{1}{9}

t a n 3 θ \approx 3 θ

= 3 \times \frac{1}{9} = 0.33

s o t a n 3 θ = 0.33 (a p p r o a x)

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